Problem: $\begin{aligned} &H(x)=10^{x} \\\\ &h(x)=H'(x) \end{aligned}$ $\int_{-3}^{2} h(x)\,dx=$
Solution: $h$ is the derivative of $H$, which means $H$ is an antiderivative of $h$. Since we know the antiderivative of $h$, we can use the fundamental theorem of calculus: For every function $h$ and its antiderivative $H$, $\int_a^b h(x)\,dx=H(b)-H(a)$. $\begin{aligned} &\phantom{=}\int_{-3}^{2} h(x)\,dx \\\\ &=H({2})-H({-3}) \\\\ &=\left[10^{{2}}\right]-\left[10^{{-3}}\right] \\\\ &=100-0.001 \\\\ &=99.999 \end{aligned}$ In conclusion, $\int_{-3}^{2} h(x)\,dx=99.999$